here is the question httpiimgurcomvnrLfV2jpg If the pitchin

here is the question : http://i.imgur.com/vnrLfV2.jpg

If the pitching wedge the golfer is using gives the ball an initial angle Theta_0 = 50 , what range of velocities v_0 will cause the ball to land within 3 ft of the hole (Assume the hole lies in the plane of the ball\'s trajectory).

Solution

For a motion of a body thrown at an angle of from horizontal with u being the magnitude of initial velocity. we have: y = x tan – (gx² sec²)2u²

Here, y and x denotes the distance along the vertical and distance along the horizontal respectively.

Now, for the above situation, we have y = 3ft

and x is to lie between 27 and 33 ft [For the ball to land within 3 feet of the hole]

Subsituting y = 3ft and x = 27, we get

3 = 27 tan50 - 315.63 x 729 x sec50 * sec50 / 2*u*u

278445.89 / U*U = 29.177

That is u = 97.68 ft/sec

Again, we substitute y = 3 and x = 33

3= 33 tan50 - 315.63*1089*sec50*sec50/2*u*u

415950.04 / u*u= 36.33

That is, u = 107 ft/sec

Therefore, for the ball to land withing 3 ft of the hole, the initital speed of it must be withing 97.68 ft/sec and 107 ft/sec

NOTE: In case you are not aware of the relation I have used above, I will quickly explain the derivation

for the horizontal direction = X = Vcos *t

while y = vsin*t - g*t*t/2

Just substitute the value of t fromt he first equation into the second, you would get the equation I have used. You can use the same for any problem with the motion in two dimension, as in one plane. This, in a way, simplifies the solution.